Using the Ratio Test

Use the Ratio Test to determine whether each series converges or diverges.

1. \(\sum\limits_{k\,=\,1}^{\infty }\frac{k}{4^{k}}\)
2. \(\sum\limits_{k\,=\,1}^{\infty }\frac{2^{k}}{k}\)
3. \(\sum\limits_{k\,=\,1}^{\infty }\frac{3k+1}{k^{2}}\)
4. \(\sum\limits_{k\,=\,1}^{\infty }\frac{1}{k!}\)

Solution (a) \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{k}{4^{k}}\) is a series of nonzero terms; \(a_{n+1}=\dfrac{n+1}{4^{n+1}}\) and \(a_{n}= \dfrac{n}{4^{n}}\). The absolute value of their ratio is \[ \left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\dfrac{\dfrac{n+1}{4^{n+1}}}{ \dfrac{n}{4^{n}}}=\dfrac{n+1}{4^{n+1}}\cdot \dfrac{4^{n}}{n}=\frac{n+1}{4n} \]

Then \[ \lim\limits_{n\,\rightarrow \,\infty }\,\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{n+1}{4n}=\frac{1}{4} <1 \]

Since the limit is less than \(1\), the series converges.

(b) \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{2^{k}}{k}\) is a series of nonzero terms; \(a_{n+1}=\dfrac{2^{n+1}}{n+1}\) and \(a_{n}=\dfrac{ 2^{n}}{n}.\) The absolute value of their ratio is \[ \left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\frac{2^{n+1}}{n+1} \cdot \frac{ n}{2^{n}} =\frac{2n}{n+1} \]

Then \[ \lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{2n}{n+1}=2>1 \]

Since the limit is greater than \(1\), the series diverges.

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(c) \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3k+1}{k^{2}}\) is a series of nonzero terms; \(a_{n+1}=\dfrac{3(n+1) +1}{(n+1) ^{2}}=\dfrac{3n+4}{(n+1) ^{2}}\) and \(a_{n}=\dfrac{3n+1 }{n^{2}}\). The absolute value of their ratio is \[ \left\vert \frac{a_{n+1}}{a_{n}}\right\vert {=}\left[ \frac{3n+4}{(n+1)^{2}} \right] \left( \frac{n^{2}}{3n+1}\right) {=}\left( \frac{3n+4}{3n+1}\right) \left( \frac{n^{2}}{n^{2}+2n+1}\right) =\dfrac{3n^{3}+4n^{2}}{3n^{3}+7n^{2}+5n+1} \]

Then \[ \lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\dfrac{3n^{3}+4n^{2}}{3n^{3}+7n^{2}+5n+1}=1 \]

The Ratio Test gives no information about this series. Another test must be used. \(\bigg({\rm{You}}\, {\rm{can}}\, {\rm{show}}\, {\rm{that}}\, {\rm{the}}\, {\rm{series}}\, {\rm{diverges}}\, {\rm{by}}\, {\rm{comparing}}\, {\rm{it}}\, {\rm{to}}\, {\rm{the}}\, {\rm{harmonic}}\, {\rm{series}}\, \sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}.\bigg)\)

(d) \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}\) is a series of nonzero terms; \(a_{n+1}=\dfrac{1}{(n+1) !}\) and \(a_{n}=\dfrac{ 1}{n!}\). The absolute value of their ratio is \[ \left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\frac{n!}{(n+1)!} = \frac{1}{n+1} \]

Then \[ \lim_{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim_{n\rightarrow \infty }\frac{1}{n+1}=0 \]

Since the limit is less than \(1,\) the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k!}\) converges.